We know that the scaling limit of a simple symmetric random walk (SSRW) $X$ on a line is a standard Brownian motion $B$. For Brownian motion $B$, and for $a \leq x \leq b$, it is easy to find the probability $$P(T_a < T_b \ \vert \ B_0 = x) = \frac{b-x}{b-a},$$ where $$T_y = \inf\{ t \geq 0 : B_t = y\}.$$ In particular, we have that $$P(T_n < T_0 \ \vert \ B_0 = 1) = \frac{1}{n}.$$ Now, if we consider the same probability but for the discrete SSRW, we know that it also $1/n$, after doing some simple computations.
I was wondering, if there is a way to extract that probability for the SSRW using only the results for the Brownian motion? In other words, is there a way to connect $P(T_n < T_0 \ \vert \ B_0 = 1)$ and $P(T_n < T_0 \ \vert \ X_0 = 1)$? Or in general, is there a way to extract some probabilities for the discrete process from its continuous scaling limit?
Well, I guess you can do the following thing:
Let $T_n^k$ the $k$th return time of the Browninan motion at $n\in \mathbb{N}$, that is, inductively, for $k\geq -1$, $$ T_n^k=\inf\{t\geq T_n^{k-1} \mid B_t=n\}, $$ with the convention that $T_n^{-1}=0$. Also, set $$T^k = \inf_{n\in \mathbb{N}}\{T_n^k \},$$ which in fact is the $k$th return time of the Browninan motion in the set $\mathbb{N}$. Now, we define $$ X_k=B_{T^k}, \forall k\geq 0. $$ Because of the symmetries of the Brownian motion, $X$ will certainly be a simple symmetric random walk. In particular, the first hitting time of $n$ for $X$ is the same as for $B$. However, I'm afraid that this doesn't fully answer your question, as it seems very specifically linked to Brownian motion. I guess some generalization can be build upon this, but not something very general.